arXiv:1706.06583v1 [math.CO] 20 Jun 2017
On the metric dimension of affine planes, biaffine planes and generalized quadrangles
Daniele Bartoli, Tam´as H´eger, Gy¨orgy Kiss, Marcella Tak´ats
Abstract
In this paper the metric dimension of (the incidence graphs of) particular partial linear spaces is considered. We prove that the metric dimension of an affine plane of order q ≥13 is 3q−4 and describe all resolving sets of that size if q ≥23. The metric dimension of a biaffine plane (also called a flag-type elliptic semiplane) of orderq ≥4 is shown to fall between 2q−2 and 3q−6, while for Desarguesian biaffine planes the lower bound is improved to 8q/3−7 underq≥7, and to 3q−9√qunder certain stronger restrictions onq. We determine the metric dimension of generalized quadrangles of order (s,1), s arbitrary. We derive that the metric dimension of generalized quadrangles of order (q, q),q ≥2, is at least max{6q−27,4q−7}, while for the classical generalized quadranglesW(q) and Q(4, q) it is at most 8q.
Keywords: resolving set, metric dimension, affine plane, biaffine plane, elliptic semiplane, generalized quadrangle
Mathematics Subject Classifications: 05C12, 05B25
1 Introduction
In this work we study the metric dimension of the incidence graph of specific finite
point-line incidence geometries.
For a connected graph G = (V, E) and x, y ∈ V, d(x, y) denotes the distance of x and y (that is, the length of the shortest path joining x and y).
Definition 1.1. Let G = (V, E) be a connected graph. A vertex v ∈ V is resolved by S = {v1, . . . , vn} ⊂ V if the ordered list (d(v, v1), d(v, v2), . . . , d(v, vn)) is unique. S is a resolving set for G if it resolves all the elements of V. The metric dimension µ(G) of G is the size of a smallest resolving set for it. A metric basis of G is a resolving set for G of size µ(G).
Resolving sets of graphs have been studied since the mid ’70s, and a lot of study has been carried out in distance-regular graphs. For more information about resolving sets and
related topics we refer to [1, 2, 3] and the references therein. The current study has been motivated by the work of Bailey and can be regarded as a continuation of [11], where the metric dimension of projective planes of order q ≥23 is determined. This result was extended for q≥13 in the recent thesis [14], so we have the following.
Theorem 1.2 ([11, 14]). The metric dimension of any projective plane of order q ≥ 13 is 4q−4.
Moreover, [11, Figure 3] lists all metric bases for projective planes of order q ≥23.
In Section 2, we use Theorem 1.2 to deduce that the metric dimension of an arbitrary affine plane of orderq ≥13 is 3q−4, and we describe all metric bases if q≥23. In Section 3, we study the metric dimension of biaffine planes (elliptic semiplanes of flag type) of order q, which turns out to be between 2q −2 and 3q− 6. We show that the metric dimension of many Desarguesian biaffine planes is 3q−o(q), and prove a general lower bound 8q/3−7 if q ≥ 7. Moreover, we provide some considerations supporting that, unlike in case of projective and affine planes, the metric dimension of a biaffine plane does rely on its finer structure, thus its exact value cannot be derived from the incidence axioms of biaffine planes only. Projective planes can be considered as generalized n-gons with n = 3. These structures were introduced by Tits in 1959. For a brief description of generalized n-gons we refer to [13, Chapter 13]. In the graph theoretical point of view, the incidence graph of a generalized n-gon is a bipartite graph with diametern and girth 2n. For n = 4, these objects are called generalized quadrangles (GQ for short). Section 4 is devoted to resolving sets of generalized quadrangles. The main result of the section is that a GQ of order (q, q) has metric dimension at least max{6q−27,4q−7}, while the metric dimension of the classical GQs W(q) and Q(4, q) is at most 8q. Note that (the incidence graphs of) projective, affine and biaffine planes of orderq have roughly 2q2 vertices, while GQs of order (q, q) have roughly 2q3.
1.1 Notation and preliminaries
In the sequel, S = PS ∪ LS always denotes a set of vertices in the incidence graph of a given partial linear space Π (in particular, an affine, biaffine or projective plane, or a generalized quadrangle), where PS is a point-set and LS is a line-set of Π. When we use graph theoretical notions in the context of a partial linear space, it should always be interpreted in the corresponding incidence graph. If a point P is incident with a line ℓ, we say that P blocks ℓ and that ℓ covers P. A blocking set of a partial linear space is a set of points that blocks every line; dually, a covering set is a set of lines that covers every point. A point P is essential for a blocking set B if B \ {P} is not a blocking set.
We apply the same notation as in [11]; in particular,P Q indicates the lines through two pointsP and Q; [P] and [ℓ] denote the set of all the lines through P and all the points on the line ℓ, respectively; once S =PS∪ LS is given, inner points and inner lines indicate points or lines in S, whereas outer points and outer lines refer to points and lines not in S; a line ℓ is skew, tangent or a t-secant to S if [ℓ]∩ PS is empty, just one point or has exactlyt elements, respectively; a point is covered if it lies on at least one line of LS,
uncovered otherwise, and it is t-covered if it lies on exactly t lines of LS. If a vertex is in S then it is trivially resolved by S. Also, the following lemma [11, Lemma 6] clearly holds for arbitrary partial linear spaces.
Lemma 1.3 ([11]). A line ℓ which intersects PS in at least two points is resolved by S. If a point lies on at least two inner lines then it is resolved by S.
We define a biaffine plane of order q as an affine plane of order q with a parallel class of lines removed. Biaffine planes are also called flag-type elliptic semiplanes. For further information, we refer the reader to [10, Section 7.4, in particular point 13]. Let Π be a biaffine or an affine plane of orderq. Then Π can be uniquely embedded into a projective plane of order q which we will denote by Π. Let ℓ∞ be the unique line in Π that has no points in Π. We call ℓ∞ the line at infinity or ideal line. The direction of a line of Π is its intersection withℓ∞ in Π; thus the points ofℓ∞ correspond to the parallel classes of Π and thus will be called directions. In the sequel, when working with an affine or biaffine plane Π, we consider it as embedded in Π and use the respective notation without further mention. If Π is a biaffine plane then a parallel class of lines, say, the class of vertical lines, is missing from Π; in Π, we denote the corresponding direction on ℓ∞ by (∞), but (∞) is not considered as a direction for the biaffine plane; and we call the sets of q pairwise non-adjacent points in Π (corresponding to the point-sets of the q vertical lines of Π) non-adjacency classes. For the sake of completeness, we recall the basic combinatorial properties of biaffine planes.
A biaffine plane of order q has q2 points and q2 lines; each point is incident with q lines, each line is incident with q points; for a non-incident point-line pair (P, ℓ), there exists exactly one line through P not intersecting ℓ, and there is exactly one point Q on ℓ not collinear withP. There areq parallel classes andqnon-adjacency classes, each containing q elements (lines or points, respectively) and partitioning the line set and the point set of the plane, respectively. For a point P or a line ℓ, C(P) and C(ℓ) will denote the non-adjacency or parallel class containing P or ℓ.
Let S = PS ∪ LS be a vertex set of the affine or biaffine plane Π. If LS contains a line with direction d then we call d a covered direction (with respect to S). A direction not covered byS is called anuncovered direction. If Π is a biaffine plane then by a blocked or unblocked class (with respect to S) we mean a non-adjacency class that contains at least one or no point of PS, respectively.
Finite generalized quadrangles can be defined in the following alternative way.
Definition 1.4. Let s and t be positive integers.
A point-line incidence geometry G = (P,L,I)
is a generalized quadrangle of order (s, t) if it satisfies the following axioms.
(GQ1) Each point is incident with t+ 1 lines and two distinct points are incident with at most one line.
(GQ2) Each line is incident with s+ 1 points and two distinct lines are incident with at most one point.
(GQ3) If (P, ℓ)⊂ P × L is a non-incident point-line pair then there is a unique pair (P′, ℓ′)⊂ P × L for which
P Iℓ′IP′Iℓ.
From this definition it is easy to derive the basic combinatorial properties of GQs.
Let G be a generalized quadrangle of order (s, t). Then
• each point is collinear with (t+ 1)s other points and each line is concurrent with (t+ 1)s other lines;
• G containsv = (s+ 1)(st+ 1) points and b= (t+ 1)(st+ 1) lines;
• if P and R are two non-collinear points of G then there are t+ 1 points in G which are collinear with both P and R;
• if e and f are two non-intersecting lines of G then there are s+ 1 lines in G which intersect both e and f.
2 Resolving sets for affine planes
It is easy to see that for a projective plane Π, a set S is a resolving set if and only if for any two distinct outer lines ℓ and ℓ′, PS ∩[ℓ] 6=PS ∩[ℓ′], and for any two distinct outer points P and P′,LS∩[P]6=LS∩[P′]. Hence the next proposition is straightforward.
Proposition 2.1 ([11], Proposition 7). S = PS ∪ LS is a resolving set for a finite projective plane if and only if the following properties hold for S:
(P1) There is at most one outer line skew to PS.
(P2) Through every inner point there is at most one outer line tangent to PS. (P1’) There is at most one outer point not covered by LS.
(P2’) On every inner line there is at most one outer point that is 1-covered by LS. The main difference between projective and affine planes is the existence of parallel lines.
The distance between two lines in an affine plane can be either 2 or 4, depending on whether they intersect or not. This leads to the following modification of Proposition 2.1;
the proof is left for the reader.
Proposition 2.2. A set S =PS∪ LS is a resolving set for an affine plane Π if and only if the following hold.
(A1) There is at most one uncovered outer point.
(A2) On every inner line, there is at most one 1-covered outer point.
(A1’) For each covered direction d, there is at most one outer skew line with direction d.
There is at most one outer skew line having an uncovered direction.
(A2’) For each inner point, there is at most one tangent line having uncovered direction.
We remark that the tangents in Proposition 2.2 (A2’) are necessarily outer and, in par- ticular, all tangent lines with a covered direction are resolved. Furthermore, if there is at most one uncovered direction then (A2’) is automatically satisfied, and (A1’) simplifies to ‘there is at most one outer skew line in each direction’.
Proposition2.3. LetS =PS∪ LS be a resolving set for the affine planeΠ, and suppose that there is a direction P ∈ ℓ∞ that contains at least two lines of LS. Let PS = PS ∪ ([ℓ∞]\ {P}). Then S = (PS,LS) is a resolving set for Π.
Proof. Relying on Proposition 2.1, we check the four properties of resolving sets for pro- jective planes for (PS,LS).
(P1): A skew line toPS intersectsℓ∞ inP. As P is a covered direction, there is at most one outer skew line to PS through P by (A1’).
(P1’): On ℓ∞, the only outer point is P, which is covered by LS. In Π, there is at most one outer point not covered by LS by (A1).
(P2): Let Z ∈ PS. We have to show that there is at most one outer tangent line to PS through Z. If Z ∈ PS then every line throughZ intersectsℓ∞in an inner point ofS with the only exception ZP. If Z ∈ℓ∞ is a covered direction then a tangent toPS through Z is skew to PS. By (A1’), there is at most one outer skew line to PS through Z. Suppose now that Z ∈ ℓ∞ is an uncovered direction. By (A1’), there is at most one outer skew line to PS intersecting ℓ∞ in an uncovered direction, so we are done.
(P2’): Let ℓ ∈ LS. We have to show that there is at most one outer point on ℓ that is 1-covered by LS. By (A2), there is at most one outer 1-covered point on ℓ in Π. The points of ℓ∞ are all inner points except P, which is covered by at least two lines ofLS.
Proposition 2.4. Let S =PS∪ LS be a resolving set for an arbitrary affine plane Π of order q. If|S| ≤3q−4 then |LS| ≥2q−3.
Proof. Lettbe the number of 1-covered outer points. Thent≤ |LS|by (A2). Using (A1) and double counting on the size of Γ = {(ℓ, P) |ℓ ∈ LS, ℓ ∈ [P],|[P]∩ LS| ≥2}, we see that
|LS|q−t≥ |Γ| ≥2(q2−1−t− |PS|).
Then
|LS|(q−1)≥2(q2−1−t−|PS|)+t−|LS|= 2(q2−1)−2(|LS|+|PS|)−t+|LS| ≥2q2−6q+6, hence
|LS| ≥ 2q2−6q+ 4
q−1 = 2q−4 + 2
q−1 >2q−4.
Theorem 2.5. Let Π be an arbitrary affine plane of order q ≥ 13. Then the metric dimension of Π is 3q−4.
Proof. LetS =PS∪LSbe a resolving set for Π of size at most 3q−4. Then, by Proposition 2.4, |LS| ≥2q−3. As there are q+ 1 directions and q >4 implies 2q−3> q+ 1, we see that there is a parallel class that contains at least two lines from LS. Let P be a point of ℓ∞ that is covered by at least two lines of LS. Thus Proposition 2.3 can be applied to see thatS is a resolving set for Π of size at most 4q−4; moreover, S
containsqcollinear points (onℓ∞). Thus, as the metric dimension of Π is 4q−4 (Theorem 1.2), we see that|S|= 4q−4, whence |S|= 3q−4 follows.
Theorem 2.6. Let Π be an arbitrary affine plane of order q ≥ 23. Then there are four types of metric bases for Π, listed in Figure 1.
Proof. In [11], the complete list1 of resolving sets of size 4q−4 in projective planes of order q ≥ 23 with |PS| ≤ |LS| is given. Thus to find the candidates for S in the list, we need to look for resolving sets with either q collinear points, or with q concurrent lines and |LS|< |PS| (the duals of the latter examples contain q collinear points but, as they have more points than lines, they are not listed).
Looking through the list of [11], we see that S is either (C3), (C5), (C29) with Z = P, (C30) withZ =Q, or the dual of (C1) or (C2). Removing the line with q collinear points and its inner points, (C5) is the only one which results in a set that is not a resolving set for Π (it violates (A2)), while (C29) with Z = P and (C30) with Z = Q give the same construction. Thus we end up with four different resolving sets depicted in Figure 1.
1Figure 3 of [11] contains a few potentially unclear details; an upgraded version is now available attached to [11] or at arXiv:1207.5469.
PSfrag replacements
R R′
e
ℓ1
ℓ0
Figure 1: The four types of smallest resolving sets for affine planes of order q≥23.
Take an arbitrary parallel class [P∞], say, the vertical lines, and choose two lines from it,eandℓ0. LetR, R′ be two arbitrary points one, and lete6=ℓ1∈[R]. Each smallest resolving set contains the following type of structure: ([e]\ {R, R′})∪([P∞]\ {e, ℓ0})∪([R]\ {e, ℓ1}).
To obtain a resolving set seen in the figure, we have to add one of the following four elements, respectively:
ℓ1; the lineℓ∈[R′] parallel with ℓ1;R′; ℓ0∩ℓ1.
3 On the metric dimension of biaffine planes
We denote an arbitrary biaffine plane of orderq by Bq, while BG(2, q) stands for the De- sarguesian biaffine plane; that is, the biaffine plane obtained from the Desarguesian affine plane AG(2, q). Note that ifq≤8 then the only biaffine plane of orderq is BG(2, q) (this follows from the well-known analogous fact for projective planes). The metric dimen- sion of biaffine planes of small order, together with other small distance regular graphs, was determined by Bailey [1]. It turned out that µ(BG(2,3)) = 4, µ(BG(2,4)) = 6 and µ(BG(2,5)) = 9. In Bailey’s work, BG(2,7) is exceptional as it is the only distance transi- tive graph on at most 100 vertices and valency between 5 and 13 whose metric dimension could not be calculated.
3.1 General case
First, we present some bounds using purely combinatorial tools, thus these results are valid for all biaffine planes.
3.1.1 Bounds on µ(Bq)
In case of biaffine planes, Proposition 2.1 needs further modifications because there are non-collinear points, too. The straightforward proof is omitted again.
Proposition 3.1. S =PS∪ LS is a resolving set forBq if and only if the following hold.
(B1) For each blocked class C, there is at most one uncovered outer point in C; further- more, there is at most one outer uncovered point in the union of unblocked classes.
(B2) On each inner line, there is at most one 1-covered point lying in an unblocked class.
(B1’) For each covered directiond, there is at most one skew outer line with direction d;
furthermore, there is at most one outer skew line having an uncovered direction.
(B2’) On each inner point, there is at most one tangent line with uncovered direction.
Note that a 1-covered point in (B2) or a tangent line in (B2’) is necessarily outer (as it is in an unblocked class, or has an uncovered direction). In particular, all 1-covered points lying in a blocked class and all tangent lines with a covered direction are resolved. Furthermore, if there is at most one uncovered direction then (B2’) is automatically satisfied, and (B1’) simplifies to ‘there is at most one outer skew line in each direction’ and, dually, if there is at most one unblocked class then (B2) is automatically satisfied and (B1) simplifies to
‘there is at most one outer uncovered point in each non-adjacency class’.
Proposition 3.2. If q ≥4 then µ(Bq)≤3q−6.
Proof. We give a construction of a resolving set of size 3qPSfrag replacements −6.
R
P1 ℓ1
T
W Z
Pq
X ℓ∞
ℓq
Figure 2: A resolving set for Bq of size 3q−6. We requireZPq∩C(P1)∈T. Let ℓ1 = {P1, . . . , Pq}, [P1] = {ℓ1, . . . , ℓq}, X = ℓ1∩ℓ∞, Z = ℓq∩ℓ∞, W = C(Pq)∩ℓq. Let T ⊂ C(P1) be a non-empty set of at most q−3 points such thatZPq∩C(P1)∈ T, XW ∩C(P1) ∈/ T (if ZPq ∩XW ∈ C(P1), we may choose another numbering for the points of ℓ1), and let R be an arbitrary point in C(P1)\(T ∪ {P1}) not covered byXW. LetPS =T∪ {P2, . . . , Pq−1}andLS ={ℓ2, . . . , ℓq−1} ∪ {XQ: Q∈C(P1)\(T∪ {P1, R})}. Then |PS|+|LS| = 3q−6. We check the requirements of Proposition 3.1 to show that S =PS∪ LS is a resolving set for Bq.
Clearly, C(Pq) is the only unblocked class and Z is the only uncovered direction; hence, (B2) and (B2’) hold trivially. Through the directionZ, the only outer skew line is ℓq (as ZPq is blocked by a point ofT); throughX, the only outer skew line isXR; through any other direction D the only possible outer skew line is DPq. Therefore, (B1’) is satisfied.
Dually, in C(Pq), the only outer uncovered point is Pq (as ℓq ∩C(Pq) = W is covered);
inC(P1), the only outer uncovered point is R; in any other class C(Pi), the only possible outer uncovered point is C(Pi)∩ℓq. Hence (B1) is also satisfied.
Note that in the above construction, the number of points may vary fromq−1 to 2q−5.
Also note that Bailey’s computations [1] show that µ(BG(2, q)) = 3q−6 forq = 4 and 5.
We proceed with some investigations on how small a resolving set for a biaffine plane may be. We will not obtain sharp results for the general case; in the Desarguesian case, stronger results will be obtained. As in case of affine resolving sets, one can prove a lower bound on the metric dimension of biaffine planes using that of projective planes; however, it is easy to obtain a lower bound directly.
Proposition 3.3. Let S be a resolving set for Bq. Then |PS| ≥ q − |S|/(q−1) and
|LS| ≥q− |S|/(q−1).
Proof. The number of skew lines toPS is at most |LS|+q=|S| − |PS|+q by (B1’). As k points in Bq block at most kq lines (with equality if and only if no two of them are collinear), we have|PS|q≥q2−(|S| − |PS|+q), that is, (q−1)|PS| ≥q2−q− |S|, whence
|PS| ≥q− |S|/(q−1) follows. Duality gives the statement forLS.
Proposition 3.4. For any biaffine plane Bq of order q, we have µ(Bq)≥2q−2.
Proof. LetS =PS∪LSbe a resolving set forBq. Suppose to the contrary that|S| ≤2q−3.
Then Proposition 3.3 gives|PS| ≥q−(2q−3)/(q−1)> q−2, hence|PS| ≥q−1. Similarly,
|LS| ≥q−1, thus |S| ≥2q−2, a contradiction.
Next we give more detailed results on the sizes of PS and LS in terms of two parameters.
These will not give a better lower bound immediately but we will make use of them later.
Definition 3.5. Let S =PS∪ LS be a resolving set for Bq. Then u =u(S) denotes the number of uncovered directions and c=c(S) denotes the number of unblocked classes.
Proposition3.6. LetS =PS∪ LS be a resolving set forBq. Then the numberδ of skew lines to PS is at most |LS|+q; moreover, δ≤ |LS|+q−u+ 1.
Proof. The lines of LS may be skew. Regarding outer skew lines, Proposition 3.1 (B1’) gives the required bound immediately.
Proposition 3.7. Let S =PS∪ LS be a resolving set for Bq. Then
|PS| ≥ u
u+ 12(q−1), (3.1)
|LS| ≥ c
c+ 12(q−1). (3.2)
Proof. Let P be an uncovered direction, ti(P) be the number of i-secant lines to PS through P, andt≥2 :=t2+t3+· · ·+tq. Then t0+t1+t≥2 =q. By Proposition 3.1 (B1’), t0 ≤1. So we gett≥2 ≥q−1−t1, hence|PS| ≥t1+2t≥2 ≥t1+2(q−1−t1) = 2(q−1)−t1.
Thust1 ≥2(q−1)−|PS|, so on theuuncovered directions we see at leastu(2(q−1)−|PS|) tangents to PS.
On the other hand, on a point of PS, by Proposition 3.1 (B2’), there can be only one tangent with an uncovered direction; hence the total number of tangents with uncovered directions is at most |PS|. These two give |PS| ≥ u(2(q −1)− |PS|), that is, |PS| ≥
u
u+12(q−1). Dually, |LS| ≥ c+1c 2(q−1).
Note that by duality, we may always assume that |PS| ≤ |LS|.
Lemma 3.8. Let S =PS∪ LS be a resolving set for Bq and suppose that |PS| ≤ |LS|. If
|S|<3(q−1) then u≤2. If|S|< 83(q−1) then u≤1; if |S|< 83(q−1)−1 then c≤4.
Proof. By Proposition 3.7, |S| ≥ 2|PS| ≥ u+1u 4(q − 1), where the right-hand side is increasing in u. If u≥3 then|S| ≥3(q−1) follows; if u≥2 then |S| ≥ 83(q−1) holds.
By Proposition 3.3, |S| < 3(q−1) implies |PS| ≥ q −2, hence c+1c 2(q−1) ≤ |LS| ≤
|S| −q+ 2, thus|S| ≥ c+12c + 1
(q−1)−1. Therefore, if|S|< 83(q−1)−1 then 53 > c+12c follows, a contradiction if c≥5.
3.1.2 Sharpness of the bounds
It is natural to ask whether the upper bound 3q−6 is sharp. It might be the case for Desarguesian biaffine planes; besides Bailey’s computations forq = 4 and 5, we will show some results later which support this possibility. However, for general biaffine planes, we provide an imaginary construction which shows that one should be careful to think that Proposition 3.2 is sharp in general. To this end, we need the following notation. For a finite plane Π, let τ(Π) denote the size of the smallest blocking set in Π. Given a point P and a line ℓ in a projective plane Πq, let Πq\[ℓ] and Πq\[P] denote the affine plane and the dual affine plane obtained by deleting ℓ and the points of ℓ, or P and the lines of P from Πq, respectively; and let (Πq\[P])∗ be the dual of Πq\[P] (which is an affine plane).
Suppose now that B is a blocking set in Πq \ [ℓ] and C is a covering set in Πq \ [P], and assume that P ∈ ℓ. Then Bℓ,P := Πq \([ℓ] ∪ [P]) is a biaffine plane in which B ∪ C is a resolving set (as there are no uncovered directions, nor unblocked classes, it is easy to verify this). Moreover, it is also easy to see that for any point Q ∈ B and any line r ∈ C, (B \ {Q})∪(C \ {r}) is also a resolving set for Bℓ,P; hence µ(Bℓ,P) ≤ τ((Πq\[P])∗) +τ(Πq\[ℓ])−2 follows.
Regarding the explicit size of this quasi-construction, the best lower bound known for the size of a blocking set in a general affine plane of order q is q+√q+ 1 [4, 7], but its sharpness is wide open; the recent paper [9] shows that there are affine planes of order q containing a blocking set of size at most 4q/3 + 5√q/3. Thus the authors suspect that the
metric dimension of some specific non-Desarguesian biaffine planes may be significantly smaller than 3q. This construction idea does not work for Desarguesian biaffine planes as τ(AG(2, q)) = 2q−1 (Jamison [12], Brouwer–Schrijver [6]).
Also, if we denote by τqa the size of the smallest blocking set an affine plane of order q may have, a general lower bound on µ(Bq) larger than 2τqa−2 would imply that for a projective plane Πq, P ∈Πq and ℓ ∈Πq, P ∈ ℓ, bothτ(Πq\[ℓ]) and τ(Πq\[P]∗) cannot be close to τqa. This would be a rather interesting phenomenon.
3.2 The metric dimension of Desarguesian biaffine planes
Now we turn our attention to Desarguesian biaffine planes where much better lower bounds than Proposition 3.4 can be obtained. As there cannot be too many lines in Πqnot blocked by PS, PS is almost a blocking set in Πq. In such a situation one may apply stability results on blocking sets, which essentially say that if there are not too many skew lines to a point-set X then X can be extended to a blocking set by adding a few points to it.
This motivates the following definitions.
Definition 3.9. A point-set X of Πq is k-extendable if X can be extended to a blocking set ofΠq by addingk points of Πq to it. A set K ofk points is a k-extender of X if X ∪ K is a blocking set. The setX is k-puncturedif it is k-extendable, but not(k−1)-extendable.
Definition 3.10. For a point P ∈Πq and a point-set X, let the index of P with respect to X, indX(P), be the number of skew lines throughP to X.
The following result is, in fact, equivalent with the formerly mentioned lower bound of Jamison and Brouwer–Schrijver for blocking sets in AG(2, q).
Result 3.11 (Blokhuis–Brouwer [5]). Let B be a blocking set of PG(2, q). Then each essential point of B is incident with at least 2q+ 1− |B| tangents to B.
Lemma 3.12. Let K be a k-extender of a k-punctured point-set X ⊂ PG(2, q). If P ∈ PG(2, q)\ K then indX(P)≤k, and ifP ∈ K then indX(P)≥2q− |X | −k+ 1.
Proof. Let X∗ = X ∪ K. Clearly, every skew line to X contains a point of K, and each point of K is essential for X∗. By Result 3.11, there are at least 2q+ 1− |X∗| = 2q− |X | −(k−1) tangents toX∗ through them. As these lines are all skew toX, we have indX(P)≥2q− |X | −k+ 1 for all P ∈ K. Furthermore, for any pointP ∈PG(2, q)\ X∗, at most k lines of [P] can be blocked by K; hence indX(P) ≤ k. Finally, for a point P ∈ X, indX(P) = 0.
Now suppose thatS =PS∪ LS is a resolving set for BG(2, q); by PG(2, q) we denote the ambient Desarguesian projective plane.
Proposition 3.13. Suppose that S = PS ∪ LS is a resolving set for BG(2, q), |S| ≤ 3q−(k+u+ 3) and c <2q− |PS| −k. Then PS ⊂PG(2, q) is not k-extendable.
Proof. Suppose to the contrary thatKis ak-extender ofPS. We may assume (by choosing the smallest appropriate k) that PS is k-punctured. As PS ∩ℓ∞ = ∅, there must be a point P ∈ ℓ∞∩K. Then, by Lemma 3.12, indPS(P) ≥ 2q− |PS| − k+ 1. If P = (∞) then this means that the number c of unblocked classes with respect to S is at least 2q− |PS| −k, a contradiction. If P 6= (∞) then there are at least 2q− |PS| −k biaffine skew lines to PS through P, among which there can be only one not in LS (Proposition 3.1 (B1’)). As there are q −1 directions different from P in BG(2, q), u of which are uncovered, |LS| ≥ (2q− |PS| −k−1) + (q−1−u) = 3q− |PS| −k−u−2 follows, so
|S|=|PS|+|LS| ≥3q−(k+u+ 2), a contradiction.
Next we proceed by showing that if S is small enough and|PS| ≤ |LS| then PS is indeed k-extendable for a suitable value of k which, under certain conditions, will lead to lower bounds on the size of a biaffine resolving set. To show extendability, we rely on stability results on blocking sets of PG(2, q).
Result 3.14 (Sz˝onyi–Weiner [16]). Let B be a set of points in PG(2, q), q = p prime, with at most 32(q+ 1)−ε points. Suppose that the numberδ of skew lines to B is less than
2
3(ε+ 1)2
/2. Then there is a line that contains at least q− q+12δ points of B.
As a line is a blocking set of PG(2, q) of size q + 1, the above theroem claims that B is j
2δ q+1
k+ 1
-extendable. Note that if we set ε = 32(q + 1) − |B|, the assumption δ < 23(ε+ 1)2
/2 can be rephrased as|B|< 32(q+ 1−√
2δ) + 1.
Result 3.15 (Sz˝onyi–Weiner [17]). Let B be a set of points in PG(2, q), q =ph, h≥ 2.
Denote the number of skew lines to B by δ and suppose that δ ≤ 1001 pq. Assume that
|B|< 32(q+ 1−√
2δ). Then B can be extended to a blocking set by adding at most δ
2q+ 1− |B| + 1 100 points to it.
The next theorem shows that µ(BG(2, q)) ≤ cq is not true in general for any constant c <3. Note that this means that there is no ‘generic’ construction (that is, a construction relying on the axioms of biaffine planes only) of this size.
Theorem 3.16. Suppose that S = PS ∪ LS is a resolving set for BG(2, q), q = ph, p prime. Assume that (i) h = 1 and q = p ≥ 17, or (ii) h ≥ 2 and p ≥ 400. Then
|S|>3q−9√q.
Proof. Suppose to the contrary that|S| ≤3q−9√q. ConsiderS embedded into PG(2, q);
thus, in the sequel, we have to take into account the non-adjacency classes and the line at infinity as lines. Note that as PS 6= ∅, c ≤ q −1. Propositions 3.3 and 3.6 yield
|PS| ≥q−2 and δ≤ |LS|+q+c+ 1 =|S| − |PS|+ 2q ≤ |S|+q+ 2<4q.
By duality, we may assume |PS| ≤ |LS|, and hence |PS| ≤ |S|/2 < 32q−4√q < 32q−1.
|S|/2 < 32(q+ 1−p
2|S|+ 2q+ 4) follows from the indirect assumption, hence |PS| <
3
2(q+ 1−√
2δ). In case (ii), 1001 pq ≥ 4q > δ also holds, thus we may use Results 3.14 and 3.15 to deduce that PS is k-extendable with k =j
2δ q+1
k+ 1
in case (i), and with k =j
δ
2q+1−|PS|+1001 k
in case (ii). In both cases, k≤8.
By Lemma 3.8, u ≤ 2, thus |S| ≤ 3q −13 ≤ 3q −(k +u+ 3) holds. Then (3.2) of Proposition 3.7 and q >16 give
c
c+ 12q−2< c
c+ 12(q−1)≤ |LS|=|S| − |PS| ≤2q−9√
q+ 2 <2q−8√ q−2, thus
1− 1
c+ 1 ≤1− 4
√q,
soc≤√q/4−1. As|PS| ≤ 32q−92√q andk ≤8,√q/4−1≤2q− |PS| −8≤2q− |PS| −k follows. Therefore, by Proposition 3.13, PS is not k-extendable, a contradiction.
Let us remark that the above proof could give a slightly better estimate than |S| >
3q−9√q but, as we do not think the result sharp, we decided to use simpler formulas.
To give a more general but considerably weaker lower bound on µ(BG(2, q)), we need the following result conjectured by Metsch and proved by Sz˝onyi and Weiner.
Result 3.17 (Sz˝onyi–Weiner [15, Theorem 4.1]). Let B be a point set inPG(2, q). Pick a point P not from B and assume that through P there pass exactlyr lines meetingB (that is containing at least 1 point of B). Then the total number of lines meeting B is at most
1 +rq+ (|B| −r)(q+ 1−r).
It is convenient to give an equivalent formulation of Result 3.17.
Result 3.18. Let δ denote the number of skew lines to a point set B in PG(2, q). Then for any point P /∈ B,
indB(P)2−(2q+ 1− |B|)indB(P) +δ ≥0. (3.3) This quadratic inequality means that the index of a point is either small or large, which will be essential for the next proof.
Proposition 3.19. Assume that|S|< 83q−7 and|PS| ≤ |LS|. Then PS is3-extendable.
Proof. Lemma 3.8 yields u≤1 and c≤4.
As in the previous proof, we use|PS| ≥q−2,δ ≤ |LS|+q+c+ 1 =|S|−|PS|+q+c+ 1≤
|S| + 7 and |PS| ≤ |S|/2 ≤ 4q/3− 72. Apply Result 3.18 with B = PS, substitute indB(P) = 4 into (3.3) and use|PS| ≤ |S|/2 andδ ≤ |S|+ 7 to obtain
0≤16−4(2q+ 1− |PS|) +δ ≤19−8q+ 3|S| ≤ −2,
a contradiction. Thus neither 4 nor 2q− |PS| −3 can be the index of a point of PG(2, q).
Thus for every pointP ∈PG(2, q), indPS(P)≤3 or indPS(P)≥2q− |PS| −2.
Let K be the set of points with large index and |K| =k. Let ℓ be a skew line to PS. If we had maxP∈ℓ{indPS(P)}=m≤ 3 thenδ ≤1 + (m−1)(q+ 1) = (m−1)q+m and, by Result 3.18 and |PS| ≤ 43q− 72,
0≤m2 −(2q+ 1− |PS|)m+ (m−1)q+m≤m2− 2
3q+9 2
m+ (m−1)q+m
=m 3 −1
q+m2 −7 2m <0
followed, a contradiction. Thus every skew line contains a point with index greater than 3, thus at least 2q− |PS| −2. This means that PS∪ K is a blocking set for PG(2, q). Now it remains to checkk ≤3. Suppose to the contrary that there exist four points with large index. When considering the ith point of these, we see at least (2q− |PS| −2)−(i−1) skew lines to PS not incident with the first i−1 points (i = 1,2,3,4), and thus 8q/3 >
|S|+ 7≥δ ≥4(2q− |PS| −2)−6≥8q/3, a contradiction.
Theorem 3.20. The metric dimension of BG(2, q) is at least 8q/3−7.
Proof. If q≤ 7 then 2q−2≥8q/3−7, thus Proposition 3.4 gives the result. Let q≥ 8.
Suppose to the contrary that there is a resolving set S = PS ∪ LS for BG(2, q) of size
|S|<8q/3−7. By duality we may assume that |PS| ≤ |LS|. By Proposition 3.19, PS is 3-extendable.
By Lemma 3.8,u≤1 andc≤4. Letk= 3. Then|S| ≤3q−(k+u+3) immediately follows fromu≤1 and|S|<8q/3−7, andc <2q−|PS|−k follows from|PS|<4q/3−7/2,c≤4 and q≥8. Thus we may apply Proposition 3.13 to obtain that PS is not 3-extendable, a contradiction.
4 On the metric dimension of generalized quadran- gles
Generalized quadrangles are well-known and much studied objects in finite geometry, see [13] as for a comprehensive book in the topic. In this section we consider two particular
types of quadrangles. A generalized quadrangle of order (s, t) is denoted by GQ(s, t).
4.1 Resolving sets for GQ(s, 1)
The simplest generalized quadrangles have order (s,1), or dually, (1, t). These objects are calledgrids, because the points of the unique GQ(s,1) form an (s+ 1)×(s+ 1) grid and the lines belong to two distinct parallel classes, each of them of size s+ 1 (corresponding to the rows and the coloumns of the grid).
In general, the grid graph Gn,m has an n by m grid as its vertex set where two vertices are adjacent if and only if they are in the same row or coloumn. (Grid graphs are also called rook graphs.) Clearly, the problem of finding a subset consisting only of points in GQ(s,1) resolving all the remaining points is equivalent to finding a resolving set for Gs+1,s+1. This latter question has been addressed in [8, Theorem 6.1] in a more general way.
Theorem 4.1 (C´aceres et al. [8]). Let Gn,m be an n×m grid, with n ≥ m ≥ 1. The metric dimension of Gn,m is given by
µ(Gn,m) =
( j2(n+m−1) 3
k, if m ≤n ≤2m−1,
n−1, if n ≥2m. (4.1)
Note that µ(Gn,m)≥j
2(n+m−1) 3
k always holds. Using this theorem we can easily deduce the metric dimension of GQ(s,1).
Corollary 4.2. The metric dimension of GQ(s,1)is ϕ(s), with
ϕ(s) =
4r+ 1, if s= 3r, 4r+ 2, if s= 3r+ 1, 4r+ 3, if s= 3r+ 2.
Proof. First of all note that ϕ(s) = µ(Gs+1,s+1) + 1 if s ≡ 0 (mod 3) and ϕ(s) = µ(Gs+1,s+1) otherwise. Let us construct the following resolving sets of GQ(s,1) to prove µ(GQ(s,1)) ≤ ϕ(s). We identify the points of GQ(s,1) with ordered pairs (i, j), 1 ≤ i, j ≤ s+ 1, and the lines with ha = {(i, a) : i ∈ {1, . . . , s+ 1}} and va = {(a, i) : i ∈ {1, . . . , s+ 1}}.
Lets = 3r+t with t ∈ {0,1,2}, and consider
S ={(1 + 3i,2 + 3i),(2 + 3i,1 + 3i),(1 + 3i,3 + 3i),(3 + 3i,1 + 3i) i= 0, . . . , r−1}. We define S′ in the following way.
1. If t = 0 then S′ :=S ∪ {vs+1}.
2. If t = 1 then S′ :=S ∪ {(1 + 3r,1 + 3r),(2 + 3r,1 + 3r)}.
3. If t = 2 then S′ :=S ∪ {(1 + 3r,2 + 3r),(1 + 3r,3 + 3r)} ∪ {vs+1}.
It is easy to see that S′ is a resolving set for GQ(s,1) of size 4r+t+ 1, so we omit the proof.
For a lower bound, suppose that a resolving setS of GQ(s,1) containsα lines of the first parallel class and β lines of the second one. The points of GQ(s,1) not contained in the α+βlines form a (s+1−α)×(s+1−β) grid. In order to resolve this set of points, we need at least µ(Gs+1−α,s+1−β) ≥ j
2(s+1−α+s+1−β−1) 3
k = j
4s+2−2(α+β) 3
k points, by Theorem 4.1.
Thus |S| ≥j
4s+2−2(α+β) 3
k+α+β, which is easily seen to be smaller thanϕ(s) if and only if α+β = 0. Thus the assertions hold if α+β > 0. Ifα+β = 0 then S consists entirely of points, thus corresponds to a resolving set for Gs+1,s+1, and so|S| ≥µ(Gs+1,s+1). This shows that for t = 1,2, |S| ≥ϕ(s). Concerning the case t = 0, note that S resolves also the lines of GQ(s,1), so it cannot have two skew lines. Following the proof of [8, Theorem 6.1], if s= 3r, all resolving sets forGs+1,s+1 of size µ(Gs+1,s+1) =ϕ(s)−1 have two skew lines, and thus do not resolve the lines of GQ(s,1). Therefore, |S| ≥ ϕ(s) holds in this case as well.
4.2 Resolving sets for GQ(q, q)
In this subsection an important class of generalized quadrangles is considered. This class contains two of the five classical quadrangles. We give a general lower bound first. If two points P and Qare adjacent, we write P ∼Q and P 6∼Qotherwise.
Proposition4.3. The metric dimension of anyGQ(q, q)is at leastmax{6q−27,4q−7}. Proof. Let Π be a GQ(q, q), S = PS ∪ LS be a resolving set for Π, PN denote the set of the uncovered outer points and let |PS| = k and |LS| = m. By duality, we may assume that m ≥ k, and hence |S| ≥ 2k. As LS covers at most m(q+ 1) points,
|PN| ≥q3+q2+q+1−k−m(q+1). The distance between any point ofPN and any line of LS is 3, so the elements ofPN must be distinguished by their distances from the elements of PS. The distance of two points is either 2 or 4, according to whether the points are collinear or not. Let Ci be the set of points of PN that are collinear with exactly i inner points. As |PS| = k, it follows from the pigeonhole principle that |Ci| ≤ ki
. First, let Γ :={(P, Q) : P ∈ PN \(C0∪C1), Q∈ PS, P ∼ Q}, and let us count |Γ| in two different ways. On the one hand, we may choose P arbitrarily from PN \(C0∪C1), and then we find at least two admissible pairs for P; thus
|Γ| ≥ q3+q2+q+ 1−k−m(q+ 1)− |C0| − |C1|
·2
≥2(q3+q2+q)−2m(q+ 1)−2k−2|C1|. (4.2)
On the other hand, each point Q ∈ PS is collinear with at most (q+ 1)q points of PN. Summing up these for all Q ∈ PS, we obtain k(q+ 1)q, but we have also counted the points of Ci exactly i times, hence we get
|Γ| ≤k(q+ 1)q− |C1|. (4.3)
Combining (4.2) with (4.3) and using |C1| ≤k and m=|S| −k we obtain k(q2−q+ 1) + 2|S|(q+ 1)−2(q3+q2+q)≥0.
Using |S| ≥2k, one may deduce
|S| ≥ 4q(q2+q+ 1)
q2+ 3q+ 5 = 4q−7− q2−5q−35
q2+ 3q+ 5 >4q−8.
Next we modify and refine the above calculations to derive |S| ≥ 6q −27. Let now Γ :={(P, Q) : P ∈ PN \(C0∪C1∪C2), Q∈ PS, P ∼Q}. Similarly as we obtained (4.2), we get
|Γ| ≥ q3+q2+q+ 1−k−m(q+ 1)− |C0| − |C1| − |C2|
·3.
For a point Q∈ PS, let PN(Q) be the number of points ofPN adjacent to Q, and denote by h the number of inner lines through Q. Then |PN(Q)| ≤(q+ 1−h)q. If h ≥1 then
|PN(Q)| ≤ q2. If h = 0 then each line of LS contains a point adjacent to Q, and that point is covered by at most q inner lines, thus |PN(Q)| ≤ (q + 1)q −m/q. If m ≥ q2 then |S| ≥q2+ 1≥6q−27, so we are done; thus we may assume m < q2, in which case
|PN(Q)| ≤ (q+ 1)q−m/q holds. Summing up these estimates for all points of PS we get |Γ| ≤k(q+ 1)q−km/q. In this estimate we have counted each point of Ci exactly i times, thus
|Γ| ≤k(q+ 1)q−km/q− |C1| −2|C2| also holds. Rearranging, substituting m = |S| − k and |Ci| ≤ ki
, the two estimates together give
1 2+ 1
q
k2− k|S| q +
q2−2q+3 2
k+ 3|S|(q+ 1)−3(q3+q2+q)≥0.
Suppose now to the contrary that |S| ≤ 6q−28. Then, using 2k ≤ |S| ≤ 6q−28 we obtain
f(q, k) :=
1 2 − 1
q
k2+
q2−2q+3 2
k+ 3(6q−28)(q+ 1)−3(q3+q2+q)≥0.
Recall that 0 ≤k ≤ |S|/2. Thus, to get a contradiction, it is enough to see that f(q,0) and f(q,3q−14) are both negative by the convexity off(q, k) in k. Asf(q,0) =−3q3+
15q2−69q−84<0 and f(q,3q−14) = −q2/2−175q/2−196/q+ 77 <0, we conclude that |S| ≥6q−27 must hold.
There are two known infinite series of classical generalized quadrangles of order (q, q). To make the paper self-contained, we recall their brief descriptions.
In PG(3, q), all points of the space and the self-conjugate lines of a null polarity with the incidence inherited from PG(3, q) form a generalized quadrangle of order (q, q).
This quadrangle is denoted by W(q). The lines of W(q) are lines in PG(3, q), hence each of them contains q+ 1 points, so axiom (GQ2) is satisfied with s=q.
Ifαis a null polarity of PG(3, q) then each point is autoconjugate. LetP be an arbitrary point. If ℓ is a line through the points P and R then ℓα is the intersection of the planes Pα and Rα. Hence ℓ is self-conjugate if and only if P ∈Rα (and R∈Pα).
As P ∈Pα also holds, this means that the self-conjugate lines through P are the lines of the pencil with carrier P in the plane Pα. Each pencil contains q+ 1 lines hence W(q) satisfies axiom (GQ1) with t=q.
Let (P, ℓ) be a non-incident point-line pair in W(q). Then ℓ is a self-conjugate line of α but it does not contain P, hence ℓ is not contained in the plane Pα.
So ℓ intersects Pα in a unique point, say P′ in PG(3, q). Then P P′ is the unique self- conjugate line through P which meets ℓ, hence axiom (GQ3) is satisfied by W(q).
LetQbe a non-singular quadric in PG(4, q). Then the points ofQand the lines contained in Q with the incidence inherited from PG(4, q) form a generalized quadrangle of order (q, q). This quadrangle is denoted byQ(4, q). It is well-known that Q(4, q) is isomorphic to the dual of W(q) and, if q is even then W(q) is self-dual. Hence, from the graph theoretic point of view, the incidence graphs ofW(q) andQ(4, q) are always isomorphic.
Definition4.4. A triple(x, y, z)of points is called a triadif no two of them are collinear.
A point u is a center of a triad if u is collinear with each of the three points of the triad.
By 1.3.6. (iii) and 3.3.1. of [13], we have the following.
Proposition 4.5. If q is odd then every triad of Q(4, q) has either 0 or 2 centers.
We proceed by constructing small resolving sets for W(q) as the union of two semi- resolving sets.
Proposition 4.6. There exists a semi-resolving set of size 4q for the points of W(q).
Proof. Embed W(q) into PG(3, q), let π be the null-polarity whose self-conjugate lines are the lines ofW(q). First we construct a semi-resolving set for the points of W(q). This construction does not depend on the parity of q. Leta1, a2 anda3 be three pairwise skew lines ofW(q). These lines define a hyperbolic quadricH in PG(3, q). Let a4 be a line of W(q) which has empty intersection with H (an easy calculations shows the existence of
such a line). We claim that the set PS = [a1]∪[a2]∪[a3]∪[a4] is a semi-resolving set of size 4q+ 4 for the points of W(q).
Let C be an outer point. Then the plane Cπ contains none of the lines a1, a2, a3 and a4, because these are autoconjugate lines. For i = 1,2,3,4 let Ai = Cπ ∩ai. As W(q) does not contain any triangle, this means that d(C, Ai) = 2 and d(C, P) = 4 for all P ∈ PS\{A1, A2, A3, A4}. First, suppose thatA1, A2 andA3 are not collinear in PG(3, q).
Then the intersection of the three planes Aπ1, Aπ2 and Aπ3 in PG(3, q) is the single point C, hence C is resolved in this case. If A1, A2 and A3 are collinear in PG(3, q) then let e denote the line joining them. Now the intersection of the three planes Aπ1,Aπ2 and Aπ3 in PG(3, q) is the line eπ. As ai ⊂ Aπ1, the lines eπ and ai are coplanar, hence eπ ∩ai 6= ∅ for i= 1,2,3. The lines a1, a2 and a3 are pairwise skew, hence H contains three distinct points of eπ, which means that the whole line eπ is contained in H. As a4 has empty intersection with
H, it has empty intersection with eπ, too. This means that if E is any point on eπ then the plane Eπ does not contain the line a4, so Eπ ∩a4 is a unique point. Hence
the points of eπ are also resolved by PS.
It is easy to see that if we delete one point from each of the lines a1, a2, a3 and a4 then the remaining 4q points also form a semi-resolving set for the points of W(q).
Corollary 4.7. If q is even then the metric dimension of W(q) is at most 8q.
Proof. If q is even then W(q) is self-dual, hence the dual of a semi-resolving set for the points is a semi-resolving set for the lines. Thus the union of the semi-resolving set constructed in the previous proposition and its dual is a resolving set for W(q) and its size is 8q.
Proposition4.8. Ifq is odd then there is a semi-resolving set of size 5q−4 for the lines of W(q), which contains exactly q−3 points, all incident with the same line.
Proof. We may look for a semi-resolving set for the points of Q(4, q) instead. Let U be an arbitrary point, and suppose that ℓ0, ℓ1, . . . , ℓq are the q+ 1 lines incident with U. Let U 6= W ∈ ℓ0, let ℓ be any line through W different from ℓ0. We claim that PS = [ℓ0]∪[ℓ1]∪[ℓ2]∪[ℓ]\ {U, W} and LS ={ℓ4, . . . , ℓq} form a semi-resolving set S for the points. U and W are clearly resolved by their distances from the points of ℓ1 and ℓ.
LetT /∈ PS,T 6=U, W.
IfT ∼U then its distances from the elements ofLS determine the unique i≥3 for which T ∈ℓi. For allR ∈ℓi, there is a unique point onℓcollinear withR, and, sincei6= 0, these points are pairwise distinct. Thus T is resolved by S. Suppose T 6∼ U. Note that there are q3 such points. As for i= 0,1,2, there is a unique point on ℓi collinear with T, so T is the center of a unique triad (x0, xi, x2) withxi ∈ℓi. Since there are q3 such triads, each having as a center U and thus, by Proposition 4.5, another point not collinear with U,
we see that any point T 6∼ U is resolved by S, which finishes the proof. The calculation of the size is easy.
Corollary 4.9. If q is odd then the metric dimension of W(q) is at most 8q−1.
Proof. Take a semi-resolving set for the lines as in Proposition 4.8, and also one for the points as in Proposition 4.6 in such a way that q−3 collinear points of the former one is contained in the latter. Then the union of these two sets is a resolving set of size at most 8q−1.
We summarize the results of this section in the following theorem.
Theorem 4.10. The metric dimension of W(q) satisfies the inequalities max{6q−27,4q−7} ≤µ(W(q))≤8q.
Acknowledgement
The research was partially supported by the Italian MIUR (progetto 40% “Strutture Geometriche, Combinatoria e loro Applicazioni”), GNSAGA, the bilateral Slovenian- Hungarian Joint Research Project no. NN 114614 (in Hungary) and N1-0032 (in Slovenia), and the Bolyai Research Grant.
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Daniele Bartoli
University of Perugia, Italy
Department of Mathematics and Informatics via Vanvitelli 1, 06123, Perugia
e-mail: daniele.bartoli@unipg.it
Tam´as H´eger, Marcella Tak´ats
MTA–ELTE Geometric and Algebraic Combinatorics Research Group ELTE E¨otv¨os Lor´and University, Budapest, Hungary
Department of Computer Science
1117 Budapest, P´azm´any P. stny. 1/C, Hungary e-mail: hetamas@cs.elte.hu,takats@cs.elte.hu Gy¨orgy Kiss
Department of Geometry and
MTA–ELTE Geometric and Algebraic Combinatorics Research Group ELTE E¨otv¨os Lor´and University, Budapest, Hungary
1117 Budapest, P´azm´any P´eter s´et´any 1/C, Hungary, and FAMNIT, University of Primorska,
6000 Koper, Glagoljaˇska 8, Slovenia e-mail: kissgy@cs.elte.hu
